# Probabilities that are "missing" in the normal distribution table

I have some difficulties using Normal tables, this is the table that I'm using.

I have the following example:

$$\begin{array}{lcl}P(Z > 1.377) & = & Q(1.377) \\ & = & 0.0842 \end{array}\\$$ How can I find the value $$Q(1.377)$$ from that table?
Looking at that table I can see only $$Q(1.37) = 0.0853$$

I have tried, but, without success to considered this:
$$Q(1.38) + \int_{1.377}^{1.38} \frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}z^2} dx \quad \mbox{ with} \quad -\infty < z < \infty$$

or also:

$$Q(1.37) - \int_{1.377}^{1.37} \frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}z^2} dx \quad \mbox{ with} \quad -\infty < z < \infty$$

please, can you help me? Thanks!

• For many students of math the first experience with interpolation from tables is either with logarithms or the (cumulative) normal distribution. One thing to keep in mind is that you are asking for a one-side tail of the normal distribution. With most tables the steps are small enough that linear interpolation is adequate. Sep 4, 2017 at 18:06

You can approximate it as $Q(1.38)+0.3(Q(1.37)-Q(1.38))$, where the $3$ comes because you went 30% of the way from 1.38 to 1.37. This linear interpolation technique is usually how students are taught to do it in elementary statistics.
• @JB-Franco Using linear interpolation is pretty normal for doing calculations with a table. It takes this particular form (interpolating between two values 0.01 apart) because of this particular table, however. For instance if the table entries were spaced 0.1 apart then the linear interpolation for this one would be $Q(1.4)+0.23(Q(1.3)-Q(1.4))$ instead.